Physics - SHM Question with Solution | TestHub

$x={\sin }^2\omega t-2{\cos }^2\omega t$

$=1-3{\cos }^2\omega t$

$=1-3\left(\frac{1+\cos 2\omega t}{2}\right)$

$= -\frac{1}{2}-\frac{3}{2}\cos 2\omega t$

Which is a periodic function with angular frequency of$2\omega$.